6 votes
Accepted

The fundamental group of $\Pi_{n=1}^\infty S^1$ with the $\operatorname{sup}$-metric

Suppose $f:S^1\to X$ is a loop. By compactness, $f$ is uniformly continuous, and it follows that the winding numbers of the coordinates of $f$ are bounded (if $\delta>0$ is such that $|x-y|<\...
Eric Wofsey's user avatar
4 votes
Accepted

Morphism inducing isomorphism in homotopy category is weak equivalence

We might have a different version of the book (mine is a 2009 ''reprint of the 1999 edition''), because I don't have a Lemma 4.1 to begin with. I do have Proposition 1.14, which states: "Suppose ...
Daniël Apol's user avatar
  • 4,132
4 votes
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Is $X\times \{0,1\}$ in the unreduced suspension contractible?

To answer the question in the title: as long as $X$ is nonempty, the space $X \times {0,1}$ is not path-connected, so it is not contractible.
John Palmieri's user avatar
3 votes

Is Minkowski space contractible?

The contractibility of a space depends only on its topology. The underlying topology on Minkowski space is the usual 'Euclidean' topology on $\mathbb{R}^4$, and this topological space is contractible ...
Jordan Payette's user avatar
3 votes
Accepted

Is Minkowski space contractible?

Don't get fooled by the Minkowski space tag. Minkowski space ist just $\mathbb{R}^4$ with a certain choice of metric. This metric, however, is not important for the Poincare lemma, since it only ...
F. Conrad's user avatar
  • 4,360
3 votes

Fundamental group of the n-fold dunce cap

Munkres is using his Theorem 72.1 here (in turn based on the Seifert-van Kampen theorem), which describes exactly how to compute the fundamental group. Indeed, it tells us that $\pi_1(A,a)\rightarrow\...
Thorgott's user avatar
  • 11.8k
3 votes
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Attaching a copy of $S^1$ at each point of $[0, 1]$ and $S^1$ and the fundamental groups of the resulting spaces

The space $B$ is indeed the wedge sum of uncountably many circles, hence its fundamental group is the free group on uncountably many generators, represented by the wedge summands. The space $C$ is ...
Thorgott's user avatar
  • 11.8k
2 votes

Seifert-van Kampen theorem, classical version

It tells you: $j_1:U\subset X$ and $j_2:V\subset X$ induce maps on $\pi_1$, $\pi_1(U)\to\pi_1(X)$ and $\pi_1(V)\to\pi_1(X)$, and these are genuine homomorphisms, and if you have two groups $A,B$ and a ...
FShrike's user avatar
  • 40.5k
2 votes

Uniqueness in Seifert-van Kampen theorem in Munkres' Topology

This is the obvious consequence of the property that $\Phi \circ j_k = \phi_k$ for $k = 1, 2$. $\pi_1(X,x_0)$ is generated by the set $\Gamma = j_1(\pi_1(U,x_0)) \cup j_2(\pi_1(V,x_0))$. For each $g_1 ...
Kritiker der Elche's user avatar
2 votes

The fundamental group of n dimensional sphere

Let's represent $S^n$ as the set of all $(x_1,x_2,\dots,x_n,x_{n+1}) \in \mathbb{R}^{n+1}$ satisfying $x_1^2 + x_2^2 + \dots + x_n^2 + x_{n+1}^2 = 1$. Then $S^n - \{p,q\}$ consists of the points in $S^...
Matthew Leingang's user avatar
2 votes
Accepted

Filtration of wedge of spheres

Let $C$ be the cone on $\mathbb{R}P^2$; then you should be able to find a filtration of that which yields torsion in homology. Since $C$ is contractible, $C \vee S^3 \vee S^3$ has the homotopy type of ...
John Palmieri's user avatar
2 votes
Accepted

Homotopy in the space of linear isometries

Let's take $u \in \Bbb{R}^\infty$ and compute (with respecto to the usual metric / Euclidian inner product / norm) : $$ \begin{align} \|(t\alpha + (1-t)\operatorname{id}_{\Bbb{R}^\infty})(u)\|^2 &...
Abezhiko's user avatar
  • 8,353
1 vote
Accepted

show that $P_* \pi_1(\tilde{X}, \tilde{x}_0)$ is exactly the given normal group

This will be too long to give in a comment so I will write it here. Throughout, let $X=S^1 \vee S^1$. N is a normal subgroup of $\pi_1(X, x_0)$ and the aim of the question is to show that $p_*(\pi_1(\...
JD1874's user avatar
  • 427
1 vote
Accepted

Point-set level object/category

In the context of homotopy theory the language of $\infty$-categories is used frequently. Building it up in forms of models founded on the theory of ordinary sets and their points is quite elaborate, ...
Jonas Linssen's user avatar
1 vote
Accepted

Prove that if a smooth manifold $M$ is contractible then every vector bundle over $M$ is trivial

You might also try Hussemoller's book "Fiber Bundles", Chapter 2, Corollary 4.8. Usually one proves, in the same breath, that if $f,g : X \to Y$ are two homotopic maps and if $B$ is a vector ...
Lee Mosher's user avatar
  • 121k
1 vote

Space homotopy equivalent to its own suspension is contractible

remark: you don't need any of this. Making a few assumptions rather than abstract isomorphisms might help. Suppose that $X$ is connected and a cw-complex. Also suppose that the natural map $X \to CX/X=...
Andres Mejia's user avatar
1 vote
Accepted

Equivalence between the category $\text{Ho}(\mathcal{M})$ and $\text{Ho}(\mathcal{M}_{\mathcal{F}})$

I don't think there is a substantial difference in complexity between showing full and faithfulness and showing an inverse functor exists. Note that a morphism $f:X\rightarrow Y$ gives rise to a ...
Jonas Linssen's user avatar

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